5.5 Black body radiation — Practise

Work through these on your own. Most of the questions want an explanation, so answer in full sentences and name the radiation involved.

Both the questions and the answers below come from a set that is checked by computer every time the book is built.

Explain why every object emits infrared radiation, even when it is cold.
A black surface and a shiny surface are at the same temperature. Compare the radiation each surface emits and absorbs.
Explain what happens to the temperature of an object when it absorbs more radiation than it emits.
Two objects have the same size and the same surface. One is hotter than the other. Compare the radiation they emit.
A warm object emits most of its radiation at a wavelength of 10 μm. Calculate the frequency of this radiation.
A hotter object emits most of its radiation at a wavelength of 2.0 μm. Calculate the frequency of this radiation.
A piece of iron is heated. It first glows dull red, then orange, then white. Explain the change.
One star looks red and another looks blue. State which star is hotter and explain your answer.
A cup of hot tea cools down and then stays at room temperature. Explain why its temperature becomes constant.
The Sun is much hotter than the Earth. Explain how their emitted radiation differs, and why the Earth still has a steady temperature.
A thermal camera shows a person as bright against a cold background. Explain what the camera is detecting.
Two objects emit their most intense radiation at 3.0 μm and 6.0 μm. Say which relationship you will use to compare their frequencies, then work out both frequencies and state which object is hotter.

Answers and concise solutions

1. Why every object emits infrared radiation

Every object emits electromagnetic radiation because of the motion of the particles inside it.
A warm object emits more infrared radiation than a cool object, but even a cold object emits some.

2. A comparison of the two surfaces

A black surface is a good absorber and a good emitter of infrared radiation.
A shiny surface is a poor absorber and a poor emitter, because it reflects most of the radiation.

3. What happens to the temperature

The object absorbs more energy each second than it emits, so its thermal store fills.
The extra energy raises the temperature of the object.

4. A comparison of the radiation emitted

The hotter object emits more radiation each second, so its total intensity is greater.
The most intense wavelength emitted by the hotter object is shorter than for the cooler object.

5. Frequency of the peak radiation

Answer: $f = \dfrac{c}{\lambda} = \dfrac{3.0 \times 10^8}{1.0 \times 10^{-5}} = 3.0 \times 10^{13}\,\text{Hz}$ .

6. Frequency of the peak radiation

Answer: $f = \dfrac{c}{\lambda} = \dfrac{3.0 \times 10^8}{2.0 \times 10^{-6}} = 1.5 \times 10^{14}\,\text{Hz}$ .

7. Why the glow changes colour

As the iron gets hotter it emits much more radiation at every wavelength.
The most intense wavelength becomes shorter. It moves from red, through orange, towards the shorter wavelengths of white light.

8. Which star is hotter

The blue star is hotter.
A hotter star emits its most intense radiation at a shorter wavelength. Blue light has a shorter wavelength than red light.

9. Why the temperature becomes constant

At first the tea is hotter than its surroundings, so it emits more radiation than it absorbs and it cools.
Once the tea is at room temperature it emits and absorbs energy at the same rate, so its temperature stops changing.

10. How the radiation differs

The Sun is much hotter, so it emits far more radiation and its most intense wavelength is much shorter.
The Earth absorbs radiation from the Sun and emits its own infrared radiation. Its temperature is steady when the two rates are equal.

11. What the thermal camera detects

The camera detects the infrared radiation emitted by the objects in the scene.
The person is warmer than the background, so the person emits more infrared radiation and appears brighter.

12. The two frequencies and the hotter object

Answer: Use $f = \dfrac{c}{\lambda}$ . The first object emits at $f = \dfrac{3.0 \times 10^8}{3.0 \times 10^{-6}} = 1.0 \times 10^{14}\,\text{Hz}$ . The second emits at $f = \dfrac{3.0 \times 10^8}{6.0 \times 10^{-6}} = 5.0 \times 10^{13}\,\text{Hz}$ . The first object has the higher frequency and the shorter peak wavelength, so it is the hotter object.